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Study Guides > Prealgebra

Solving Word Problems Involving Tickets and Stamps

Learning Outcomes

  • Apply the problem-solving method to solve problems involving tickets with different values
  • Apply the problem-solving method to solve problems involving stamps with different values
 

The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values, so we can organize the information in tables much like we did for coin problems.

Example

At a school concert, the total value of tickets sold was $1,506\text{\$1,506}. Student tickets sold for $6\text{\$6} each and adult tickets sold for $9\text{\$9} each. The number of adult tickets sold was 55 less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

Solution:

Step 1: Read the problem.

  • Determine the types of tickets involved.
    There are student tickets and adult tickets.
  • Create a table to organize the information.
Type Number\text{Number} Value ($)\text{Value (\$)} Total Value ($)\text{Total Value (\$)}
Student 66
Adult 99
1,5061,506

Step 2. Identify what you are looking for.

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was 55 less than three times the number of student tickets sold.

Let ss be the number of student tickets.

Then 3s−53s - 5 is the number of adult tickets.

Multiply the number times the value to get the total value of each type of ticket.

Type Number\text{Number} Value ($)\text{Value (\$)} Total Value ($)\text{Total Value (\$)}
Student ss 66 6s6s
Adult 3s−53s - 5 99 9(3s−5)9\left(3s - 5\right)
1,5061,506

Step 4. Translate: Write the equation by adding the total values of each type of ticket.

6s+9(3s−5)=15066s+9\left(3s - 5\right)=1506

Step 5. Solve the equation.

6s+27s−45=150633s−45=150633s=1551s=47 students\begin{array}{c}6s+27s - 45=1506\hfill \\ 33s - 45=1506\hfill \\ 33s=1551\hfill \\ s=47\text{ students}\hfill \end{array}

Substitute to find the number of adults.

3s−5=3s-5= number of adults 3(47)−5=1363(\color{red}{47})-5=136 adults  

Step 6. Check. There were 4747 student tickets at $6\text{\$6} each and 136136 adult tickets at $9\text{\$9} each. Is the total value $1506?\text{\$1506}? We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.

\begin{array}{ccc}\hfill 47\cdot 6& =\hfill & 282\hfill \\ \hfill 136\cdot 9& =\hfill & \underset{\text{_____}}{1224}\hfill \\ & & 1506\quad\checkmark \hfill \end{array}

Step 7. Answer the question. They sold 4747 student tickets and 136136 adult tickets.

 
 
 

Now we'll do one where we fill in the table all at once.

Example

Monica paid $10.44\text{\$10.44} for stamps she needed to mail the invitations to her sister's baby shower. The number of 49-cent\text{49-cent} stamps was four more than twice the number of 8-cent\text{8-cent} stamps. How many 49-cent\text{49-cent} stamps and how many 8-cent\text{8-cent} stamps did Monica buy?

Answer:

Solution:

The type of stamps are 49-cent\text{49-cent} stamps and 8-cent\text{8-cent} stamps. Their names also give the value.

"The number of 49-cent\text{49-cent} stamps was four more than twice the number of 8-cent\text{8-cent} stamps."

Letx=number of 8-cent stamps2x+4=number of 49-cent stamps\begin{array}{c}\text{Let}x=\text{number of 8-cent stamps}\\ 2x+4=\text{number of 49-cent stamps}\end{array}

Type Number\text{Number} Value ($)\text{Value (\$)} Total Value ($)\text{Total Value (\$)}
49-cent\text{49-cent} stamps 2x+42x+4 0.490.49 0.49(2x+4)0.49\left(2x+4\right)
8-cent\text{8-cent} stamps xx 0.080.08 0.08x0.08x
10.4410.44
Write the equation from the total values. 0.49(2x+4)+0.08x=10.440.49\left(2x+4\right)+0.08x=10.44
Solve the equation. 0.98x+1.96+0.08x=10.440.98x+1.96+0.08x=10.44
1.06x+1.96=10.441.06x+1.96=10.44
1.06x=8.481.06x=8.48
x=8x=8
Monica bought 8 eight-cent stamps.
Find the number of 49-cent stamps she bought by evaluating. 2x+4 for x=82x+4\text{ for }x=8.
2x+42x+4
2⋅8+42\cdot 8+4
16+416+4
2020
Check.
8(0.08)+20(0.49)=?10.448\left(0.08\right)+20\left(0.49\right)\stackrel{?}{=}10.44
0.64+9.80=?10.440.64+9.80\stackrel{?}{=}10.44
10.44=10.44✓10.44=10.44\quad\checkmark

Monica bought eight 8-cent\text{8-cent} stamps and twenty 49-cent\text{49-cent} stamps.

  Watch the following video to see another example of how to find the number of tickets sold given the total sales for two different ticket values. https://youtu.be/dRFp1Db6ymA

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